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in a particular game a player throws two fair dice together.
if their total is 7 or 11 he wins
if their total is 2,3 or 12 he looses
for any other total, he neither wins or looses.
(i) find the probability of winning
(ii) find the probability of losing
 11 months ago
 11 months ago
in a particular game a player throws two fair dice together. if their total is 7 or 11 he wins if their total is 2,3 or 12 he looses for any other total, he neither wins or looses. (i) find the probability of winning (ii) find the probability of losing
 11 months ago
 11 months ago

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tcarroll010Best ResponseYou've already chosen the best response.1
Start with listing all your possible dice pair outcomes and count the number of 7's and 11's out of all possible outcomes. That gives your probability of winning. Same process with the losing for 2, 3, and 12.
 11 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
So start with "all possible outcomes". How many of those are there? hint: each die has 6 faces.
 11 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
Better start with one die. How many outcomes?
 11 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
Yes. Now for EACH GIVEN outcome on that first die, you can have 6 outcomes on the second die. So, 6 x 6 = 36 possible outcomes for a roll of 2 die. Now, it really would be beneficial for you to write out on paper a 2dimensional table where horizontal is one die and vertical is the other die and make a table of 36 rolls. That will benefit you immensely here.
 11 months ago

samnathaBest ResponseYou've already chosen the best response.0
is the answer for part (i) 2over 9 ?
 11 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
Yes, 2/9 for winning. That comes from 6 7's and 2 11's inning rolls out of a possible 36: 8/36 = 2/9 Good job!
 11 months ago

samnathaBest ResponseYou've already chosen the best response.0
yeah ok thank you is is the same method for loosing aswell ?
 11 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
Very similar. Add up all 2's (there's just 1), all 3's (there are 2) and all 12's (there's 1) for that total over 36.
 11 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
Yes, because 4/36 = 1/9. This would be a nice game to play because you win more than lose!
 11 months ago

samnathaBest ResponseYou've already chosen the best response.0
thank you :) i actuallt have a similar question to that and i was wondering if i got it right or not do you mind checking for me ?
 11 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
Good luck in all of your studies and thx for the recognition! @samnatha I'll check it if you like.
 11 months ago

samnathaBest ResponseYou've already chosen the best response.0
thank you i think i sent to you in mail :)
 11 months ago

samnathaBest ResponseYou've already chosen the best response.0
i'll just put it here aswell :)
 11 months ago

samnathaBest ResponseYou've already chosen the best response.0
this circle is divided into 6 equal sectors. you pay €10 to spin the arrow and you win the amount in the sector where the arrow stops. what is the expected amount you win or loose in this game ? theres a picture of a cirlce aswell divided into 6 sections in each section there is different numbers which are €0 €30 €10 €10 €0 and €5
 11 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
Since you pay 10 to play, we are really talking about the "net" winning or losing, so you can actually redo the 6 sectors as a "net" amount: 10, +20, 0, 0, 10, 5 You add those up and divide by 6 to get 5/6. Not a good game to play. Over time, you lose big.
 11 months ago

samnathaBest ResponseYou've already chosen the best response.0
ok i'm slighlt confused on how you did all that not great at probability
 11 months ago

tcarroll010Best ResponseYou've already chosen the best response.1
np. Each sector is equally likely, and for each sector, we can conceptually see ourselves handing over 10 (at first, to play) and then getting back some money (if any for that sector). So, for example, if we landed on "get 30", it's a net of 20 because we had to pay 10 up front. Your expected value is the net amount times the probability of landing in that sector. It is 1/6 times each sector and then all added up. Or you could just add them all up first: (1/6)(10) + (1/6)(20) + (1/6)(0) + (1/6)(0) + (1/6)(10) + (1/6)(5) or what I did: (1/6)(10 + 20 + 0 + 0 + 10 5)
 11 months ago

samnathaBest ResponseYou've already chosen the best response.0
ooh ok thank you i am use to doing it the other way so that made more sense thanks alot :)
 11 months ago
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